Question

Calcium carbonate decomposes to form calcium oxide and carbon dioxide, like this:

Answer 1

Answer: The value of the equilibrium constant Kc for this reaction is 0.088

Explanation:

$Molarity=\frac{x}{M\times V_s}$

where,

x = given mass

M = molar mass

$V_s$ = volume of solution in L

Equilibrium concentration of $CaCO_3$ = $\frac{25.3}{100\times 9.0}=0.028M$

Equilibrium concentration of $CaO$ = $\frac{14.9}{56\times 9.0}=0.029M$

Equilibrium concentration of $CO_2$ = $\frac{33.7}{44\times 9.0}=0.085M$

The given balanced equilibrium reaction is,

$CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CaO]\times [CO_2]}{[CaCO_3]}$  

Now put all the given values in this expression, we get :

$K_c=\frac{0.029\times 0.085}{0.028}=0.088$